The reaction between hydrogen and fluorine is shown in the chemical equation below using displayed formulae:
H−H+F−F⟶2 H−F \text{H}-\text{H} + \text{F}-\text{F} \longrightarrow 2\ \text{H}-\text{F} H−H+F−F⟶2 H−FThe bond energies are given in the table below:
| Bond | Bond energy in kJ/mol\text{kJ/mol}kJ/mol |
|---|---|
| H−H\text{H}-\text{H}H−H | 436 |
| F−F\text{F}-\text{F}F−F | 158 |
| H−F\text{H}-\text{F}H−F | 562 |
Which expression shows how to calculate the overall energy change for this reaction?
436+158+562 kJ/mol436 + 158 + 562 \text{ kJ/mol}436+158+562 kJ/mol
436+158+(2×562) kJ/mol436 + 158 + (2 \times 562) \text{ kJ/mol}436+158+(2×562) kJ/mol
436+158−562 kJ/mol436 + 158 - 562 \text{ kJ/mol}436+158−562 kJ/mol
436+158−(2×562) kJ/mol436 + 158 - (2 \times 562) \text{ kJ/mol}436+158−(2×562) kJ/mol
13 exam-style questions on AQA GCSE Chemistry 5.1 Exothermic and endothermic reactions, covering 5.1.1 Energy transfer during exothermic and endothermic reactions, 5.1.2 Reaction profiles, and 5.1.3 The energy change of reactions (HT only). Each one has a worked solution and a mark scheme showing where the marks go.